Complex analysis antiderivative existence

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SUMMARY

The discussion confirms that the function f(z) = 1/z does not have an antiderivative over the complex plane excluding the origin, C/(0,0), due to the non-zero integral around the unit circle, which equals 2πi. Consequently, this indicates that f(z) is not path independent. Similarly, for the function f(z) = (1/z)^n, where n is an integer not equal to 1, the same reasoning applies, leading to the conclusion that it also lacks an antiderivative over the specified domain.

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This discussion is beneficial for students and professionals in mathematics, particularly those studying complex analysis, as well as educators seeking to clarify concepts related to antiderivatives and contour integration.

reb659
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Homework Statement



a) Does f(z)=1/z have an antiderivative over C/(0,0)?

b) Does f(z)=(1/z)^n have an antiderivative over C/(0,0), n integer and not equal to 1.

Homework Equations


The Attempt at a Solution



a) No. Integrating over C= the unit circle gives us 2*pi*i. So for at least one closed contour the integral is nonzero. Thus f(z) cannot be path independent and thus cannot have an antiderivative over the domain.

b) By using the same reasoning it seems that no antiderivative over the domain exists either, but I'm not sure.
 
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reb659 said:

Homework Statement



a) Does f(z)=1/z have an antiderivative over C/(0,0)?

b) Does f(z)=(1/z)^n have an antiderivative over C/(0,0), n integer and not equal to 1.

Homework Equations





The Attempt at a Solution



a) No. Integrating over C= the unit circle gives us 2*pi*i. So for at least one closed contour the integral is nonzero. Thus f(z) cannot be path independent and thus cannot have an antiderivative over the domain.

b) By using the same reasoning it seems that no antiderivative over the domain exists either, but I'm not sure.

For b) why don't you make sure by integrating over a contour? Try z=e^(i*t) for t in [0,2pi].
 

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